Trig Identities and Double Angles

Date: 01/05/98 at 11:41:52
From: Doctor Wilkinson
Subject: Re: Trig Identities- Double Angles
My first step in a problem like this would be to express everything in
terms of sines and cosines, since you have sines and cosines on the
right side.
Also I know the identities for sines and cosines pretty well, but I
tend to forget them for the other functions.
So
sec2A + tan2A = 1 + sin2A = 1 + sin2A
----- ----- ---------
cos2A cos2A cos2A
Now apply the double angle formulas for the sin and cosine to get
1 + 2sinA cosA
-------------------
cos^2(A) - sin^2(A)
Now it's time to get clever, because there are a number of directions
we might consider going. The denominator factors as a difference of
squares:
cos^2(A) - sin^2(A) = (cosA + sinA)(cosA - sinA)
This is kind of promising, because one of the factors is cosA - sinA,
which is what we want to get in the denominator. This tells us we
probably don't want to substitute
cos^2(A) = 1 - sin^2(A) or sin^2(A) = 1 - cos^2(A)
in the denominator. We already have the form we want in the
denominator; let's look at what we can do with the numerator.
At first glance things don't look so good. We could substitute for sin
in terms of cosine or cosine in terms of sine, but that would bring
square roots in, and would complicate things rather than simplify
them.
A useful trick in trig identities is sometimes to make a substitution
that looks crazy until you see how it works. Let's substitute
sin^2(A) + cos^2(A) for 1
This turns the numerator into
sin^2(A) + cos^2(A) + 2 sinAcosA
which we can arrange to
sin^2(A) + 2 sinAcosA + cos^2(A)
which factors into
(sin(A) + cos(A))^2 = (cosA + sinA)^2
and the rest is easy.
-Doctor Wilkinson, The Math Forum
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